Motion in a Straight Line - Result Question 27

28. A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

[1994]

(a) (α2+β2αβ)t

(b) (α2β2αβ)t

(c) (α+β)tαβ

(d) αβtα+β

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Answer:

Correct Answer: 28. (d)

Solution:

velocity, v=dsdt=2at3bt2

acceleration a=dvdt=2a6bt

Acceleration is zero at

2a6bt=0t=a3b

In Fig.

AA1=vmax. =αt1=βt2

But t=t1+t2=vmaxα+vmax β

or, vmax=t(αβα+β)

If a body starting from rest accelerates at a constant rate α for certain time and then retards at constant β and comes to rest after t. second from the starting point, then

Distance travelled by the body =αβt2(2α+2β)



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